JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Sequences and Series

Let the common difference be $d=\frac{100-a}{n+1}$.

The first arithmetic mean is $A_1=a+d$ and the last is $A_n=100-d$.

Given $\frac{A_1}{A_n}=\frac{1}{7}$:
$\frac{a+d}{100-d}=\frac{1}{7}$
$7a+7d=100-d$
$7a+8d=100$

Substituting $d=\frac{100-a}{n+1}$:
$7a+8\cdot \frac{100-a}{n+1}=100$

Using $a+n=33$, we get $a=33-n$. Substituting and solving:
$7n^2-132n-667=0$

Solving this quadratic: $n=23$ (the other root is negative and rejected).